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3 years ago

How do I solve this Physics problem?

Physics

1 answer:

schepotkina [342]3 years ago

3 0

Like this:

1). Ignore the 22.7 m/s horizontal speed. It doesn't make a bit of difference, and the answer doesn't depend on it. (A bullet fired horizontally from a rifle and another bullet dropped from the end of the rifle barrel at the same time both hit the ground at the same time.)

2). Calculate how long it takes an object dropped from rest to fall 42.3 m. Use the "dropped from rest" formula:

Distance = (1/2) (acceleration) (time)²

3). In the formula, "acceleration" is the acceleration of gravity. If this story happened on Earth, then use the acceleration of gravity on Earth. That's 9.81 m/s² .

I wish I could make it more complicated for you, but I don't know how.

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alukav5142 [94]

Answer:

The correct answers would be option A, C and E.

Explanation:

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3 years ago

A race car starting from rest accelerates uniformly at a rate of 4.90 meters per squared. What is the cars speed after it has tr

shusha [124]

Ok, we need to find a relation for the speed as it relates to the acceleration. This is given by the integral of acceleration:

$v= \int\limits^{t}_{0} {a} \, dt' =at$

Where we have the initial velocity is 0m/s and a will be 4.90m/s².

But we see there is an issue now... We know the velocity as a function of time, but we don't know how long the car has been accelerating! We need to calculate this time by now finding the position function as a function of time. This way we can solve for the time, t, that it takes to go 200m accelerating this way and then substitute that time into our velocity equation and get the velocity.
Position is just the integral of velocity:

$s= \int\limits^{}_{} {at} \, dt = \frac{1}{2}at^2$

Where the initial velocity and initial position are both zero.

Now we set this position function equal to 200m and find the time, t, it took to get there

$\frac{1}{2}(a \frac{m}{s^2} )t^2=200m \\ \\ \frac{1}{2}4.90 \frac{m}{s^2} t^2=200m \\ \\ t^2= \frac{400m}{4.90 \frac{m}{s^2}}=81.63s^2 \\ \\ t= \sqrt{81.63s^2 } =9.04s$

Now let's put t=9.04s into our velocity equation:

$v =at=4.9\frac{m}{s^2} \times 9.04s=44.3 \frac{m}{s}$

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4 years ago

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